Tight end will join him on the Tampa Bay Buccaneers
Four-time All Pro tight end Rob Gronkowski has agreed to a reunion with Tom Brady. The agent for the retired New England star confirmed Tuesday that pending completion of a physical, Gronkowski has agreed to play for the Tampa Bay Buccaneers, who are acquiring his rights from the Patriots.
A proposed trade that needs to be finalized before this week’s NFL draft would bring Gronkowski and a seventh-round pick to Tampa Bay in exchange for a fourth-round selection, the AP reports. “Pending the physical, Rob has agreed to play for Tampa this season,” agent Drew Rosenhaus said.
The deal would reunite the 30-year-old Gronkowski with Brady, who signed a two-year, $50 million contract with the Bucs last month. NBC News notes the two were “notably close” when they played together for the Patriots.
“Rob has played his entire career alongside Tom Brady and their accomplishments speak for themselves,” Buccaneers’ General Manager Jason Licht said in a statement. “Together they have developed the type of chemistry on and off the field that is crucial to success.”
Gronkowski, who’ll turn 31 on May 14, retired in March 2019 after nine seasons with the Patriots, who drafted him in the second round in 2010. He has one year left on his contract at $10 million. “He will honor his current contract at this time,” Rosenhaus said. In addition to 521 receptions for 7,861 yards and 79 touchdowns in 115 regular-season games, the five-time Pro Bowl selection has another 81 catches for 1,163 yards and 12 TDs in 16 playoff games.